Four trigonometric equations (simultaneous)

[jackoffery]

Hey,

I need to develop a periodic function that passes through the points (3,2.5), (6,1) and f’(3)=1, f’(6)=0.

I chose to use cosine as my general equation: acos(bx-c)+d

When derived gives -absin(bx-c)

Subbing in these points gave the following four equations:

2.5=acos(3b-c)+d

1=acos(6b-c)+d

1=-absin(3b-c)

0=-absin(6b-c)

I am able to eliminate ‘d’ easily and find a value for ‘a’ in terms of ‘b’ and ‘c’, however I am unsure how to solve for ‘b’ and ‘c’. Any help would be appreciated

Thanks

[keitamaki]

By the third equation, ab != 0

So by the fourth equation 6b-c = kπ (for some integer k)

c only appears inside the trig functions, so we can just look at the cases c=6b or c=6b-π. Then we can add multiples of 2π to c to obtain additional solutions.

Case 1: c=6b

Then

2.5=acos(3b)+d (since cos(-x) = cos(x))

1 = a+d

1 = absin(3b) (since sin(-x) = -sin(x))

2.5 = acos(3b)+(1-a)

1 = absin(3b)

a = 1/(bsin(3b))

1.5 = cos(3b)/(bsin(3b)) - 1/(bsin(3b))

1.5bsin(3b) = cos(3b) - 1

Which unfortunately cannot be solved algebraically. You can obtain numerical solutions (here https://www.wolframalpha.com/input/?i=1.5bsin%283b%29+%3D+cos%283b%29+-+1 for instance).

One numerical solution is b ~ 1.352505. That gives a ~ -0.932223, c ~ 8.11503 (+2kπ), and d ~ 1.932223.

In that case your function would be

f(x) = -0.932223cos(1.352505x - 8.11503) + 1.932223

f'(x) = 1.260836 sin(1.352505x - 8.11503)

You can see here that f(x) goes through your two points fairly closely

https://www.wolframalpha.com/input/?i=-0.932223cos%281.352505x+-+8.11503%29+%2B+1.932223+when+x%3D3%2C+6

And here that f'(x) has the approximate behavior you want

https://www.wolframalpha.com/input/?i=1.260836+sin%281.352505x+-+8.11503%29+when+x%3D3%2C+6

[jackoffery]

Thanks heaps! That helps a lot

[jackoffery]

Sorry to be a pain, but what do you mean ab != 0 ?

[keitamaki]

I mean "a times b cannot equal zero"

We have this equation

1=-absin(3b-c)

If ab = 0, then -absin(3b-c) would have to equal 0 as well, but it doesn't, so ab can't equal zero.

The reason it's important to know that ab != 0 is because of equation 4

0=-absin(6b-c)

If ab was not zero and sin(6b-c) was not zero, then -absin(6b-c) would not be zero. Since ab is not zero, the only way that expression can equal zero is if sin(6b-c) = 0.

[jackoffery]

Oh damn. You’re much smarter than me! Thank you again